Divide a uniform sample in half. Its mass halves; its temperature does not. This simple experiment separates properties that depend on system size from those that describe local state.
Describe a state
What you’ll be able to do
Distinguish intensive, extensive, and specific properties.
Separate a state property from a process quantity.
Recognize the conditions needed for equilibrium.
Before you touch the controls
Two identical samples are combined without changing their state. Which doubles: mass, volume, density, or temperature?
Explore the modelChange a control. Follow the response.
Whole sample0.004 m³
Temperature=20°C; density=1,000kg/m³ in every portion.
Total internal energy at u = 80 kJ/kg320 kJ
Internal energy in sample320 kJ
The total extensive properties add. The temperature, density, and specific internal energy do not change when you divide the sample at the same state.
01
Extensive properties scale with amount.
Mass, volume, and total internal energy are extensive. For identical portions at the same state, their totals add. Intensive properties such as temperature, pressure, and density do not scale with sample size. Dividing a uniform sample is a useful classification test.
02
Specific properties remove the mass dependence.
A specific property is an extensive property per unit mass. Specific volume v=V/m has units m³/kg; density ρ = m/V is its reciprocal. Specific internal energy u=U/m has units J/kg. Internal-energy zero points are conventional; changes are what matter in balances.
v=mV=ρ1,u=mU
03
A state is not a path.
Properties describe a state. Heat and work describe a process between states and are not stored properties. For a simple compressible system of fixed composition, two independent intensive properties determine an equilibrium state. They must actually be independent: pressure and temperature are linked during liquid-vapor coexistence.
04
Equilibrium is stronger than a constant reading.
Thermal equilibrium means no temperature differences driving heat transfer. Mechanical and chemical equilibrium impose additional restrictions. A wall carrying steady heat flow can have temperatures that are constant in time but different from place to place; steady state is not necessarily equilibrium.
A worked example
One sample, two descriptions.
A uniform liquid has m=4kg and V=0.004 m³. Its density is 1,000 kg/m³ and specific volume is 0.001 m³/kg.
Split it into equal portions without changing temperature or pressure. Each has m=2kg and V=0.002 m³.
The ratios remain unchanged. Density and specific volume are intensive even though mass and volume are extensive.
Make the idea your own
Predict. Calculate. Question the model.
Try each problem before opening its solution. Then change a value to test whether the reasoning still holds.
01 · Calculate
Scale the sample, keep the state.
A 2 kg sample occupies 0.002 m³ and has specific internal energy 80 kJ/kg. At the same state, find V, U, and density for 6 kg.
A small hint
Volume and total energy scale with mass; their mass-specific values remain fixed.
Explore the visual solution
Total internal energy480 kJ
Volume of the sample0.006 m³
Density=1,000kg/m³; u=80kJ/kg. Both stay constant.
What if you change it?
The visual responds to your control. The worked steps below use the original question.
Original question · Step 1 of 3
Specific volume is 0.002/2=0.001 m³/kg.
Carry this forward.A property can be extensive even when its corresponding specific property is intensive.
02 · Calculate
Will these temperatures average to 50°C?
Mix 1 kg of water at 80°C with 3 kg at 20°C in an insulated container. Assume constant equal specific heats, no phase change, and negligible container heat capacity. Find the equilibrium temperature.
A small hint
Energy lost by the hotter water equals energy gained by the colder water.
Explore the visual solution
Equilibrium mixture temperature35°C
Tf = (1×80+3×20)/(1+3). Equal constant specific heats; no heat loss.
What if you change it?
The visual responds to your control. The worked steps below use the original question.
Original question · Step 1 of 3
Write 1c(Tf−80) + 3c(Tf−20) = 0.
Carry this forward.The warmer sample can have a higher temperature yet contain less mass and lower total energy than the colder sample.
03 · Challenge
Steady or equilibrium?
A wall has one face maintained at 80°C and the other at 20°C. The temperature profile no longer changes with time. Is the wall in thermal equilibrium?
A small hint
A time derivative can be zero while a spatial gradient is not.
Explore the visual solution
Thermal stateSteady, not equilibrium
Temperature is constant in time. The line shows its variation across wall position; the arrow shows heat-flow direction.
What if you change it?
The visual responds to your control. The worked steps below use the original question.
Original question · Step 1 of 3
The profile is steady because each location stays at a constant temperature over time.
Carry this forward.Steady means unchanging in time; equilibrium additionally excludes the driving forces for spontaneous change.
Your learning path
Ready to carry this forward?
Mark this lesson complete when you can explain the main idea and work through its practice. Completion is your choice.
